A hydraulic cylinder on an industrial machine pushes a steel block a distance of x feet (0 ≤ x ≤ 6), where the variable force required is F(x) = 1100xe−x pounds. Find the work done in pushing the block the full 6 feet through the machine. (Round your answer to three decimal places.)

W = 1080.914 J

Explanation:

f(x) = 1100xe⁻ˣ

Work done by a variable force moving through a particular distance

W = ∫ f(x) dx (with the integral evaluated between the interval that the force moves through)

W = ∫⁶₀ 1100xe⁻ˣ dx

W = 1100 ∫⁶₀ xe⁻ˣ dx

But the integral can only be evaluated using integration by parts.

∫ xe⁻ˣ dx

∫ vdu = uv - ∫udv

v = x

(dv/dx) = 1

dv = dx

du = e⁻ˣ dx

∫ du = ∫ e⁻ˣ dx

u = -e⁻ˣ

∫ vdu = uv - ∫udv

∫ xe⁻ˣ dx = (-e⁻ˣ)(x) - ∫ (-e⁻ˣ)(dx)

= -xe⁻ˣ - e⁻ˣ = -e⁻ˣ (x + 1)

∫ xe⁻ˣ dx = -e⁻ˣ (x + 1) + C (where c = constant of integration)

W = 1100 ∫⁶₀ xe⁻ˣ dx

W = 1100 [-e⁻ˣ (x + 1)]⁶₀

W = 1100 [-e⁻⁶ (6 + 1)] - [-e⁰ (0 + 1)]

W = 1100 [-0.0173512652 + 1]

W = 1100 × (0.9826487348)

W = 1080.914 J

Hope this Helps!!!